winMySQL 查询在 Ubuntu 实例上失败，但适用于 Windows 实例

Question

为什么同一个查询在 Windows MySQL 实例上可以工作，但在 Ubuntu MySQL 实例上却不起作用？

实例版本非常接近：

• Windows：8.0.16
• Ubuntu：8.0.39-0ubuntu0.24.04.2

---

Windows： @@sql_mode：STRICT_TRANS_TABLES，NO_ENGINE_SUBSTITUTION

Ubuntu： @@sql_mode：ONLY_FULL_GROUP_BY、STRICT_TRANS_TABLES、NO_ZERO_IN_DATE、NO_ZERO_DATE、ERROR_FOR_DIVISION_BY_ZERO、NO_ENGINE_SUBSTITUTION

---

Ubuntu 上的错误：

错误 3065 (HY000): ORDER BY 子句的表达式 #1 不在 SELECT 列表中，引用列 'p.priority' 不在 SELECT 列表中；这与 DISTINCT 不兼容

---

查询：

SELECT DISTINCT 
    `id`, `date_created`, `date_last_update`, `text_id_map`, `address_search_aid`, `lat`, `lng`, 
    `street_line1`, `street_line2`, `zip`, `company_status`, `compassion_level`, `emails`, 
    `phones`, `social_media`, `urls`, `description`, `importance`, `name`, `object_status`, 
    `search_aid`, `short_name`, `created_by_id`, `last_update_by_id`, `alternate_country_id`,
    `city_id`, `parent_id`, `city_region_id`, `autocomplete`, `sitemap_xml`
FROM (
    
    SELECT p.*, 0 AS `priority`, CASE WHEN `parent_id` IS NULL THEN 1 ELSE 0 END AS `is_parent`
    FROM `provider` p 
    CROSS JOIN JSON_TABLE( JSON_KEYS(`name`), '$[*]' COLUMNS (locale VARCHAR(10) PATH '$') ) AS l 
    CROSS JOIN JSON_TABLE( JSON_EXTRACT(`name`, CONCAT('$.', l.locale)), '$' COLUMNS (`jsonval` VARCHAR(2048) PATH '$') ) AS j
    WHERE
        INSTR( CONCAT(' ', REPLACE(j.jsonval, '-', ' '), ' '), CONCAT(' ', 'web', ' ') ) > 0
        AND
        INSTR( CONCAT(' ', REPLACE(j.jsonval, '-', ' '), ' '), CONCAT(' ', 'kon', ' ') ) > 0 
    
    UNION
    
    SELECT *, 
        CASE 
            WHEN 
                INSTR( CONCAT(' ', `autocomplete`, ' '), CONCAT(' ', 'web', ' ') ) > 0 
                AND 
                INSTR( CONCAT(' ', `autocomplete`, ' '), CONCAT(' ', 'kon', ' ') ) > 0 
            THEN 1 
            WHEN INSTR( `autocomplete`, 'web' ) > 0 AND INSTR( `autocomplete`, 'kon' ) > 0 THEN 2 
            ELSE 3 
        END `priority`, 
        CASE 
            WHEN parent_id IS NULL THEN 1 
            ELSE 0 
        END AS `is_parent`
    FROM `provider`
    
) p
WHERE `priority` < 3
ORDER BY `priority`, `is_parent` DESC
LIMIT 10;

---

表是这样创建的：

CREATE TABLE `provider` (
  `id` int NOT NULL,
  `date_created` varchar(80) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `date_last_update` varchar(80) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `text_id_map` text COLLATE utf8mb4_unicode_ci,
  `address_search_aid` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `lat` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `lng` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `street_line1` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `street_line2` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `zip` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `company_status` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `compassion_level` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `emails` varchar(255) COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `phones` varchar(255) COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `social_media` text COLLATE utf8mb4_unicode_ci,
  `urls` varchar(255) COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `description` text COLLATE utf8mb4_unicode_ci,
  `importance` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `name` text COLLATE utf8mb4_unicode_ci,
  `object_status` varchar(16) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `search_aid` text COLLATE utf8mb4_unicode_ci,
  `short_name` text COLLATE utf8mb4_unicode_ci,
  `created_by_id` int DEFAULT NULL,
  `last_update_by_id` int DEFAULT NULL,
  `alternate_country_id` int DEFAULT NULL,
  `city_id` int DEFAULT NULL,
  `parent_id` int DEFAULT NULL,
  `city_region_id` int DEFAULT NULL,
  `autocomplete` text COLLATE utf8mb4_unicode_ci,
  `sitemap_xml` text CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci,
  PRIMARY KEY (`id`),
  KEY `FKotfd80u3xayhxxjrvtpclqwe6` (`created_by_id`),
  KEY `FK7gqnbcylk4ghwciki0779g32u` (`last_update_by_id`),
  KEY `FKk6jlj4kwu3hi5kl4a4qqso7j3` (`alternate_country_id`),
  KEY `FKrlh8uhcluf8si4vfbgdw3w6p1` (`city_id`),
  KEY `FK6psluxn6a0b64bcxggfuum2l0` (`parent_id`),
  KEY `FKegsn16mm766i6j5du01dlvn6` (`city_region_id`),
  CONSTRAINT `FK6psluxn6a0b64bcxggfuum2l0` FOREIGN KEY (`parent_id`) REFERENCES `provider` (`id`),
  CONSTRAINT `FK7gqnbcylk4ghwciki0779g32u` FOREIGN KEY (`last_update_by_id`) REFERENCES `veg_user` (`id`),
  CONSTRAINT `FKegsn16mm766i6j5du01dlvn6` FOREIGN KEY (`city_region_id`) REFERENCES `city_region` (`id`),
  CONSTRAINT `FKk6jlj4kwu3hi5kl4a4qqso7j3` FOREIGN KEY (`alternate_country_id`) REFERENCES `country` (`id`),
  CONSTRAINT `FKotfd80u3xayhxxjrvtpclqwe6` FOREIGN KEY (`created_by_id`) REFERENCES `veg_user` (`id`),
  CONSTRAINT `FKrlh8uhcluf8si4vfbgdw3w6p1` FOREIGN KEY (`city_id`) REFERENCES `city` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci

Windows 上 Heidi 的屏幕截图；查询从这个问题复制：

Answer 1

SQL 模式

ONLY_FULL_GROUP_BY

影响查询的执行方式。

https://dev.mysql.com/doc/refman/8.0/en/group-by-handling.html 说：

...如果任何

DISTINCT

表达式不满足以下条件中的至少一个，则具有

ORDER BY

和

ORDER BY

的查询将被拒绝为无效：

• 表达式等于选择列表中的1

• 表达式引用的所有列以及属于查询的选定表的列都是选择列表的元素

阅读我链接到的手册页，以获取有关此规则存在原因的更多说明和示例。

如果您的 SQL 模式省略

ONLY_FULL_GROUP_BY

，则此规则无效，但这会导致您的查询出现任意结果。

我建议保留

ONLY_FULL_GROUP_BY

模式。从 MySQL 5.7.5 开始默认启用。这对于 MySQL 按照标准 SQL 运行非常重要。

大多数其他品牌的 SQL 都强制执行相同的规则，但没有选项可以禁用它。